1 Answer
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Here's a hint to get started. Rewrite the equation of the given line as $y=2x+1$. This line has slope $2$, so the line you're trying to find should also have slope $2$, in order to be parallel to the given line. Also, any line tangent to $f$ has slope $2x$. If you set these two slopes equal to each other, you solve for the $x$-coordinate of a line tangent to $f$ with slope $2$, which is then parallel to the given line, as desired.

Once you have that, you can find a point on your desired line. Since you already know the needed slope, you then have a slope and a point on the line, and you can use point-slope form to write your equation.

Ok, I didn't know about setting the derivative and the slope equal to find an x for a point on the line, thanks. we need to use the point slope formula (nipissingu.ca/calculus/tutorials/lineargifs/point-slope.gif) to get the equation of the line we are looking for right? So setting the derivative and the slope equal gave us $x_1$ = 1 to use in the formula but how do we get the $y_1$ for that formula?
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MattJun 4 '11 at 1:14

@Matt, the $x$-coordinate you find is of the point at the intersection of the desired line and $f$, so the point is also on $f(x)=x^2$. You then can just plug in $x_1$ to get $y_1$.
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yunoneJun 4 '11 at 1:16

solve for $y_1$ by substituting $x_1$ into the equation $y = x^2$, then you'll have the point ($x_1, y_1)$, together with the needed slope, and can thus find the equation of the line tangent to the parabola, parallel to the given line. (just saw your post, yunone; I'm pokey with my writing answers and comments; by the time I finish, someone has often already posted!)
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amWhyJun 4 '11 at 1:18